The quadratic equation $x^2+mx+n=0$ has roots that are twice those of $x^2+px+m=0,$ and none of $m,$ $n,$ and $p$ is zero. What is the value of $n/p?$
Answer: Let $r_1$ and $r_2$ be the roots of $x^2+px+m=0.$ Since the roots of $x^2+mx+n=0$ are $2r_1$ and $2r_2,$ we have the following relationships: \[
m=r_1 r_2,\quad n=4r_1 r_2,\quad p=-(r_1+r_2), \quad\text{and}\quad
m=-2(r_1+r_2).
\] So \[
n = 4m, \quad p = \frac{1}{2}m,
\quad\text{and}\quad
\frac{n}{p}=\frac{4m}{\frac{1}{2}m}=\boxed{8}.
\]
Alternatively, the roots of \[
\left(\frac{x}{2}\right)^2 + p\left(\frac{x}{2}\right) + m = 0
\] are twice those of $x^2 + px + m = 0.$ Since the first equation is equivalent to $x^2 + 2px + 4m = 0,$ we have \[
m = 2p \quad\text{and}\quad n = 4m, \quad\text{so}\quad \frac{n}{p} = \boxed{8}.\]